Physics
PHS 5042-2 Kinematics & MomentumKinematic Equations
1st Kinematic Equation:
a = Δv / ΔtIn many cases t1 = 0, so
v2 = a t2 + v1
or just:v2 = a t + v1
aΔt = Δv
v2 – v1 = a * Δt
v2 = aΔt + v1
PHS 5042-2 Kinematics & MomentumKinematic Equations
Example:Determine the velocity of an airplane moving with constant acceleration of 5 m/s2 along a rectilinear path after 5 minutes, if its initial velocity is 20 m/s.
v2 = at + v1
v2 = (5m/s2)(300s) + 20m/sv2 = 1520m/s
PHS 5042-2 Kinematics & MomentumKinematic Equations
2nd Kinematic Equation:
Find the variation of displacement between the interval of time highlighted on the graph.
Δd = A rectangle + A triangle
Δd = (v1-0)(t2-t1) + (v2-v1)(t2-t1) / 2
Δd = v1Δt + Δv(Δt ) / 2
Δd = 10.5 m2
PHS 5042-2 Kinematics & MomentumKinematic Equations
2nd Kinematic Equation:
Δd = v1Δt + ΔvΔt/2
In many cases t1 = 0, sod2 = at2
2/2 + v1t2 + d1
or just:d2 = at2/2 + v1t + d1
Δv = aΔt
Δd = v1Δt + aΔt Δt/2
d2 = aΔt2/2 + v1Δt + d1
d2 - d1 = v1Δt + aΔt2/2
PHS 5042-2 Kinematics & MomentumKinematic Equations
Example:
A car moves 5m down a ramp with an initial velocity of 1m/s. How long before it reaches the bottom of the ramp if it moves with constant acceleration of 2m/s2?
d2 = 1/2at2 + v1t + d1
0 = ½ (-2m/s2)t2 + (-1m/s)t + 5m
0 = ½ (2m/s2)t2 + (1m/s)t - 5m
0 = (1m/s2)t2 + (1m/s)t - 5m
t = 1.79 s t = -2.79s
NEVER!
PHS 5042-2 Kinematics & MomentumKinematic Equations
3rd Kinematic Equation:v2
2 - v12 = 2aΔd
Example:A car moving at 50km/h needs 31m to come to a stop. What is the acceleration of the car as it breaks?
v22 - v1
2 = 2aΔda = v2
2 - v12 / 2Δd
a = {[0]2 - [(50 * 1000/3600)m/s]2} / 2(31m)a = - (13.9 m/s)2 / 62m
a = - 3.12 m/s2
PHS 5042-2 Kinematics & MomentumKinematic Equations
Rectilinear motion with uniform acceleration
Uniform rectilinear motion
Main characteristic a = constant v = constant
Position vs time graph
Position equation d2 = at2/2 + v1t + d1 d2 = v1t + d1 (a = 0)
Velocity vs time graph
Velocity equation v2 = a t + v1 v2 = v1 (a = 0)
Acceleration vs time graph
Other equation v22 - v1
2 = 2aΔd (a constant)
v22 - v1
2 = 0 (a = 0)
P
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P
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P
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P
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a
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V
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V
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a
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PHS 5042-2 Kinematics & MomentumKinematic Equations
Free fall
Vertical falling movement due to the pull of the Earth Uses same motion equations than rectilinear
movement with uniform acceleration Instead of “a”, we use “g” (g = - 9.8 m/s2) Instead of “d”, we use “y” (Cartesian)
PHS 5042-2 Kinematics & MomentumKinematic Equations
Free fall
Upward movement: positive direction
(displacement, velocity, acceleration) Downward movement: negative direction
(displacement, velocity, acceleration) d = 0 (y = 0) ground, d > 0 (positive, above
ground), d < 0 (negative, below ground)
PHS 5042-2 Kinematics & MomentumKinematic Equations
Free fall
y2 = gt2/2 + v1t + y1
v2 = g t + v1
v22 - v1
2 = 2gΔy
PHS 5042-2 Kinematics & MomentumKinematic Equations
Example:A key falls from a height of 70 cm. What is its velocity when it hits the ground?
A key falls from a height of 70 cm. What is the duration of the flight?
v22 - v1
2 = 2gΔyv2
2 = 2gΔy + v12
v22 = 2(-9.8 m/s2)(-0.7m) + 0
v22 = √(13.72 m2/s2)
v2 = ± 3.7 m/sv2 = - 3.7 m/s (falling)
v2 = gΔt + v1 Δt = v2 - v1 / g = v2 / g Δt = v2 / g Δt = (- 3.7 m/s / -9.8 m/s2) Δt = 0.38 s